An inclined plane is making an angle β with horizontal. A projectile is projected from the bottom of the plane with a speed u at an angle α with horizontal then the time of flight T is:
A
T=2usin(α−β)gcosβ
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B
T=usin(α−β)gcosβ
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C
T=2usin(α+β)gcosβ
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D
T=usin(α+β)gcosβ
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Solution
The correct option is BT=2usin(α−β)gcosβ Velocity perpendicular to the inclined plane uv=usin(α−β) a=−gcosβ t=T s=0 0=usin(α−β)T−12(gcosβ)T2 ⇒T=2usin(α−β)gcosβ